STRUCTURE OF CAESIUM ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( September 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks discovered by Gell-Mann and Zweig I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838,68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Caesium (Cs) has 40 known isotopes. The atomic masses of these isotopes range from 112 to 151. Only one isotope, Cs-133, is stable. The longest-lived radioisotopes are Cs-135 with a half-life of 2.3 million years, Cs-137 with a half-life of 30.1671 years and Cs-134 with a half-life of 2.0652 years. All other isotopes have half-lives less than 2 weeks, most under an hour. STRUCTURE OF Cs-123, Cs-125, Cs-127, AND Cs-129 WITH S = +1/2 After a careful analysis I found that the structure of the above unstable nuclides is based on the structure of the Cs-110 with S =0 in which the odd number of extra neutrons gives S = +1/2. For example the Cs-129 with S = + 1/2 of 19 extra neutrons has 10 extra neutrons of positive spins and 9 extra neutrons of negative spins giving S = +1/2. In other words the spin of Cs-129 is given by S = 0 + 10(+1/2) + 9(-1/2) = +1/2. These extra neutrons make two bonds per neutron but the no high symmetry under the small number of extra neutrons cannot give enough binding energies to pn bonds for overcoming the pp and nn repulsions. In this group we include the Cs-142 with S = 0, in which there are 32 extra neutrons of opposite spins giving S = 0 . In the following diagram of Cs-110 with S = 0 we clear why the additional p55 and n55 break the high symmetry of the structure of Xenon . (See my STRUCTURE OF Xe-124 ). ' ' ' DIAGRAM OF Cs-110 WITH S = 0 ' In this structure of S = 0 based on Xe-108 with 54 protons and 54 neutrons you see the 6 horizontal planes of opposite spins like the +HP1, -HP2, +HP3, -HP4, +HP5 and -HP6 along the two horizontal squares, the -HSQ , and +HSQ. Here the p55n55 as a vertical system with S=0 is not shown and breaks the high symmetry of Xenon. ' n40.......p40' ' +HSQ p38..........n38 ' ' n31………p12.........n12.......p32' ' -HP6 p31........n11.........p11…… n32 ' ' p29.........n10.........p10…… n30' ' +HP5 n29………p9..........n9 …….p30' ' p47.......n27.........p8............n8.........p28........n48' ' -HP4 n45.......p27.........n7..........p7.........n28..........p46 ' ' n47......p25.........n6.........p6..........n26...........p48' ' +HP3 p45......n25……….p5..........n5……….p26.........n46 ' ' n23……p4..........n4………….p24' ' -HP2 p23……....n3………p3………n24 ' ' p21.........n2………p2...........n22' ' +HP1 n21........p1........n1............p22' ' p37......n37 ' ' -HSQ n39.....p39' ' ' STRUCTURE OF Cs-113, Cs-115, Cs-117, Cs-119, Cs-121, Cs-131, AND Cs-133 For understanding the structure of the above nuclides you must read my STRUCTURE OF Cs-133 . Here the deuteron p37n37 changes the spin from S = -1 to S = +1 giving S = +2 , because it moves from the -HSQ to +HP5 near the n29p29. Whereas the p38n38 does not change the spin, because it moves from +HQS to +HP5 near the p30n30. Under such symmetrical arrangements the structure of the above nuclides is based on Cs-110 with S = +2 . For example the unstable Cs-131 of 21 extra neutrons with S = +5/2 has 11extra neutrons of positive spins and 10 extra neutrons of negative spins. That is, the spin of Cs-131 is given by S = +2 + 11(+1/2) + 10(-1/2) = +5/2 These extra neutrons make two bonds per neutron but the small number of extra neutrons cannot give enough binding energies to pn bonds for overcoming the pp and nn repulsions. However in the stable Cs-133 with the two more extra neutrons of positive spins than those of Cs-131 give a stable structure with S = +7/2 given by S = +2 +11(+1/2) + 10(-1/2) + 2(+1/2) = +7/2 In this case under the symmetrical arrangements the greater number of extra neutrons gives enough binding energies to pn bonds for overcoming the repulsions. ' ' STRUCTURE OF Cs-135, Cs-137,Cs-139, Cs-141, Cs-143, Cs-145, Cs-147, Cs-149, AND Cs-151 Similarly the structure of the above unstable nuclides is based on the same structure of Cs-110 with S = +2. For example the Cs-135 with S= +7/2 has two more extra neutrons of opposite spins than those of the stable Cs-133 with S = +7/2, but in the absence of blank positions they make single bonds leading to the decay. SRUCTURE OF Cs-112, Cs-114, Cs-116, Cs-118, Cs-122, Cs-124, Cs-126, Cs-128, Cs-130, AND Cs-132 After a careful analysis I found that the structure of the above unstable nuclides is based on another structure of Cs-110 having S = +1 because the vertical p55n55 with S= 0 becomes a deuteron with S = +1. In this case it goes to the +HQS for makimg horizontal bonds with p38n38. Under such arrangements the even number of extra neutrons gives the structure of the above nuclides. For example the Cs-112 with S = +1 has 2 extra neutrons of opposite spins, while the Cs-132 with S = +2 has 2 extra neutrons of positive spins and 20 extra neutrons of opposite spins giving S =0. That is S = +1 + 2(+1/2) + 0 = +2 ' ' STRUCTURE OF Cs-120, Cs-138, Cs-140, Cs-144, AND Cs-146 ''' In this group of unstable nuclides which have also even number of extra neutrons but with negative spins we conclude that they are based on a similar structure of Cs-110 having S = -1, because the vertical p55n55 becomes a deuteron with S = -1. Especially it goes to -HSQ for making horizontal bonds with p39n39. Under this arrangement the above nuclides with even number of extra neutrons are based on Cs-110 with S = -1. For example the Cs-146 with S =-1 has 36 extra neutrons with opposite spins '''STRUCTURE OF Cs-134 AND Cs-136 These cases of even number of extra neutrons are based on another structure of Cs-110 having S = +4 because the two deuterons of the -HSQ move to +HSQ giving S = +4. Under this arrangements the structures of the above nuclides are based on Cs-110 with S = +4. For example the Cs-134 with S = +4 has 24 extra neutrons of opposite spins, while the Cs-136 with S = +5 has two extra neutrons of positive spins and 24 extra neutrons of opposite spins giving S = 0 . That is S = +4 + 2(+1/2) + 0 = +5 Category:Fundamental physics concepts